Question

What is the density of atoms/nm^2 on the (110) plane of a zinc blende lattice with lattice spacing 0.546 nm. Three significant digits, fixed point notation.

Answer 1

Answer:

The density is 10.25 $\frac{atom}{nm^{2} }$

Explanation:

From the question we are given that the lattice spacing  nm

        V = $(0.546)^{3} }$ × $nm^{3}$

            = 0.1628 $nm^{3}$

           zinc blende lattice has 4 atom per unit cell.

           This means that 4  atoms is contained in 0.15056 $nm^{3}$

            Density of the cubic unit in terms of atom  is given as =

                                                                                                      =26.565 $\frac{atoms}{nm^{3} }$

            For plane  (110) the spacing between the cubic unit in the crystal denoted by  $d_{hkl}$ is given as =  $\frac{a}{\sqrt{1^{2} + 1^{2} +0^{2} } }$ Where a is the lattic spacing = 0.546

                                               =   $\frac{0.546}{\sqrt{2} }$  = 0.376 nm

               The density of the lattice in (110) plane  =26.565 $\frac{atoms}{nm^{3} }$ × 0.386 nm

                                                                                 = 10.25 $nm^{2}$